Linear Momentum for Grade 12

1. Chapter 2
Linear Momentum
 Momentum
 Conservation of Momentum
 Collisions

2. Linear Momentum
 How can the effect of catching a slow, heavy object be the same as
catching a fast, lightweight object?
 The answer: They have the same linear momentum.
 Linear Momentum 𝑷 is defined as the mass times the velocity.
𝑷 = 𝒎 × 𝒗 (𝑖𝑛 𝑆𝐼 𝑷 𝑖𝑠 𝑖𝑛 𝒌𝒈.
𝒎
𝒔
)
 Linear Momentum 𝑷 is a vector quantity, it has same direction as 𝒗
 Since Linear momentum is the product of mass and velocity, an
object's momentum changes whenever its mass or velocity changes.

3. Linear Momentum
 Click to play the video

4.  The figure shows two objects,
a beanbag bear and a rubber
ball, each with the same mass
and same downward speed just
before hitting the floor.
 What is the change in
momentum of each of the
objects?
The following example clearly illustrates
why the vector nature of momentum must
be taken into account when determining the
change in momentum of an object.

5.  If the beanbag has a mass of 1 kg and is moving
downward with a speed of 4 m/s just before
coming to rest on the floor, then its change in
momentum is
 A 1-kg rubber ball with a speed of 4 m/s just
before hitting the floor will bounce upward with
the same speed. Therefore, the ball's change in
momentum is

6. Application 1 (page 33 on the book)
𝑷 = 𝒎 × 𝒗

7. Linear Momentum of a system
of particles
 The total momentum of a system of particles is the vector sum
of the linear momenta of its particles:
 𝑷 𝒔𝒚𝒔 = 𝑷 𝟏 + 𝑷 𝟐 + 𝑷 𝟑 … . +𝑷 𝒏 = 𝑷𝒊
= 𝑚1 𝑣1 + 𝑚2 𝑣2 + … . . +𝑚 𝑛 𝑣 𝑛
 Due to the vector nature of momentum, it is possible for a
system of several moving objects to have a total momentum that
is positive, negative, or zero.

8. Linear Momentum of the
center of mass of a system
 The position vector of the center of mass of the above system is
given by:
𝑟𝐺 =
𝑚1 𝑟1 + 𝑚2 𝑟2 + … . . +𝑚 𝑛 𝑟𝑛
𝑚1 + 𝑚2 + … . . +𝑚 𝑛
 Differentiate both sides w.r.t time:
𝑣 𝐺 =
𝑚1 𝑣1 + 𝑚2 𝑣2 + … . . +𝑚 𝑛 𝑣 𝑛
𝑚1 + 𝑚2 + … . . +𝑚 𝑛
Thus, 𝑀 𝑣 𝐺 = 𝑚1 𝑣1 + 𝑚2 𝑣2 + … . . +𝑚 𝑛 𝑣 𝑛
So, 𝑷 𝑮 = 𝑷 𝒔𝒚𝒔
 Conclusion: The linear momentum of a system of particles of
constant mass is equal to the linear momentum of the center of
mass of the system.
Where m is the mass
of each particle, which
is constant, and 𝑟 is its
position vector

9. Application 2 (page 33 on the book)
𝑷 𝒔𝒚𝒔 = 𝑷 𝟏 + 𝑷 𝟐 + 𝑷 𝟑 … . +𝑷 𝒏
𝑷 𝑮 = 𝑷 𝒔𝒚𝒔

10. General Expression of
newton’s 2nd Law
 The time derivative of the linear momentum of a particle is equal to
the vector sum of the external forces acting on this particle:
𝐹𝑒𝑥𝑡 =
𝑑𝑃
𝑑𝑡
(For short duration of time
𝑑𝑃
𝑑𝑡
=
∆𝑃
∆𝑡
)
( 𝑭 𝒆𝒙𝒕 = 𝒎𝒂 𝒃𝒖𝒕 𝒂 =
𝒅𝒗
𝒅𝒕
⇒ 𝑭 𝒆𝒙𝒕 =
𝒎𝒅𝒗
𝒅𝒕
=
𝒅𝒎𝒗
𝒅𝒕
=
𝒅𝑷
𝒅𝒕
)
 The sum of the external forces acting on a system of particles is
equal to the linear momentum of its center of mass
𝑭 𝒆𝒙𝒕 =
𝒅𝑷 𝒔𝒚𝒔
𝒅𝒕
but 𝑷 𝒔𝒚𝒔 = 𝑷 𝑮 = 𝑴𝒗 𝑮 ⇒ 𝒅𝑷 𝒔𝒚𝒔 = 𝑴𝒂 𝑮
So, the center of mass theorem will be
𝑭 𝒆𝒙𝒕 = 𝑴𝒂 𝑮

11. Application 3 (page 34 on the book)